Answer:
The power developed in HP is 2702.7hp
Explanation:
Given details.
P1 = 150 lbf/in^2,
T1 = 1400°R
P2 = 14.8 lbf/in^2,
T2 = 700°R
Mass flow rate m1 = m2 = m = 11 lb/s Q = -65000 Btu/h
Using air table to obtain the values for h1 and h2 at T1 and T2
h1 at T1 = 1400°R = 342.9 Btu/h
h2 at T2 = 700°R = 167.6 Btu/h
Using;
Q - W + m(h1) - m(h2) = 0
W = Q - m (h2 -h1)
W = (-65000 Btu/h ) - 11 lb/s (167.6 - 342.9) Btu/h
W = (-65000 Btu/h ) - (-1928.3) Btu/s
W = (-65000 Btu/h ) * {1hr/(60*60)s} - (-1928.3) Btu/s
W = -18.06Btu/s + 1928.3 Btu/s
W = 1910.24Btu/s
Note; Btu/s = 1.4148532hp
W = 2702.7hp
Answer:
diesel engine
Explanation:
because diesel is stronger than petrol
Answer:
Explanation:
From the question we are told that:
Initial Pressure 
Initial Temperature 
Final Pressure 
Final Temperature 
Work Output 
Generally Specific Energy from table is
At initial state
With
Specific Volume 
At Final state
Generally the equation for The Process is mathematically given by
Assuming Mass to be Equal
Where
Therefore
Answer:
insert (array[] , value , currentsize , maxsize )
{
if maxsize <=currentsize
{
return -1
}
index = currentsize-1
while (i>=0 && array[index] > value)
{
array[index+1]=array[index]
i=i-1
}
array[i+1]=value
return 0
}
Explanation:
1: Check if array is already full, if it's full then no component may be inserted.
2: if array isn't full:
- Check parts of the array ranging from last position of range towards initial range and determine position of that initial range that is smaller than the worth to be inserted.
- Right shift every component of the array once ranging from last position up to the position larger than the position at that smaller range was known.
- assign new worth to the position that is next to the known position of initial smaller component.
